本部分举例说明了 怎么解决作业问题 MIP 求解器和 CP-SAT 求解器。
示例
示例中有五个工作器(编号为 0-4)和四个任务(编号为 0-3)。请注意,这里的 worker 比示例中的 worker 多 概览。
为任务分配工作器的费用显示在 表格。
工作器 | 任务 0 | 任务 1 | 任务 2 | 任务 3 |
---|---|---|---|---|
0 | 90 | 80 | 75 | 70 |
1 | 35 | 85 | 55 | 65 |
2 | 125 | 95 | 90 | 95 |
3 | 45 | 110 | 95 | 115 |
4 | 50 | 100 | 90 | 100 |
问题是为每个工作器分配至最多一个任务, 没有两个工作器在执行 相同的任务,同时最大限度降低总费用。 由于工作器数量超过 一个工作器将不会被分配任务。
MIP 解决方案
以下各部分介绍了如何使用 MPSolver 封装容器。
导入库
以下代码会导入所需的库。
Python
from ortools.linear_solver import pywraplp
C++
#include <memory> #include <vector> #include "ortools/base/logging.h" #include "ortools/linear_solver/linear_solver.h"
Java
import com.google.ortools.Loader; import com.google.ortools.linearsolver.MPConstraint; import com.google.ortools.linearsolver.MPObjective; import com.google.ortools.linearsolver.MPSolver; import com.google.ortools.linearsolver.MPVariable;
C#
using System; using Google.OrTools.LinearSolver;
创建数据
以下代码会为问题创建数据。
Python
costs = [ [90, 80, 75, 70], [35, 85, 55, 65], [125, 95, 90, 95], [45, 110, 95, 115], [50, 100, 90, 100], ] num_workers = len(costs) num_tasks = len(costs[0])
C++
const std::vector<std::vector<double>> costs{ {90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95}, {45, 110, 95, 115}, {50, 100, 90, 100}, }; const int num_workers = costs.size(); const int num_tasks = costs[0].size();
Java
double[][] costs = { {90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95}, {45, 110, 95, 115}, {50, 100, 90, 100}, }; int numWorkers = costs.length; int numTasks = costs[0].length;
C#
int[,] costs = { { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 }, }; int numWorkers = costs.GetLength(0); int numTasks = costs.GetLength(1);
costs
数组与费用表相对应
用于向任务分配工作器,如上所示。
声明 MIP 求解器
以下代码声明了 MIP 求解器。
Python
# Create the mip solver with the SCIP backend. solver = pywraplp.Solver.CreateSolver("SCIP") if not solver: return
C++
// Create the mip solver with the SCIP backend. std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP")); if (!solver) { LOG(WARNING) << "SCIP solver unavailable."; return; }
Java
// Create the linear solver with the SCIP backend. MPSolver solver = MPSolver.createSolver("SCIP"); if (solver == null) { System.out.println("Could not create solver SCIP"); return; }
C#
Solver solver = Solver.CreateSolver("SCIP"); if (solver is null) { return; }
创建变量
以下代码会为问题创建二进制整数变量。
Python
# x[i, j] is an array of 0-1 variables, which will be 1 # if worker i is assigned to task j. x = {} for i in range(num_workers): for j in range(num_tasks): x[i, j] = solver.IntVar(0, 1, "")
C++
// x[i][j] is an array of 0-1 variables, which will be 1 // if worker i is assigned to task j. std::vector<std::vector<const MPVariable*>> x( num_workers, std::vector<const MPVariable*>(num_tasks)); for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { x[i][j] = solver->MakeIntVar(0, 1, ""); } }
Java
// x[i][j] is an array of 0-1 variables, which will be 1 // if worker i is assigned to task j. MPVariable[][] x = new MPVariable[numWorkers][numTasks]; for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { x[i][j] = solver.makeIntVar(0, 1, ""); } }
C#
// x[i, j] is an array of 0-1 variables, which will be 1 // if worker i is assigned to task j. Variable[,] x = new Variable[numWorkers, numTasks]; for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { x[i, j] = solver.MakeIntVar(0, 1, $"worker_{i}_task_{j}"); } }
创建限制条件
以下代码为问题创建了约束条件。
Python
# Each worker is assigned to at most 1 task. for i in range(num_workers): solver.Add(solver.Sum([x[i, j] for j in range(num_tasks)]) <= 1) # Each task is assigned to exactly one worker. for j in range(num_tasks): solver.Add(solver.Sum([x[i, j] for i in range(num_workers)]) == 1)
C++
// Each worker is assigned to at most one task. for (int i = 0; i < num_workers; ++i) { LinearExpr worker_sum; for (int j = 0; j < num_tasks; ++j) { worker_sum += x[i][j]; } solver->MakeRowConstraint(worker_sum <= 1.0); } // Each task is assigned to exactly one worker. for (int j = 0; j < num_tasks; ++j) { LinearExpr task_sum; for (int i = 0; i < num_workers; ++i) { task_sum += x[i][j]; } solver->MakeRowConstraint(task_sum == 1.0); }
Java
// Each worker is assigned to at most one task. for (int i = 0; i < numWorkers; ++i) { MPConstraint constraint = solver.makeConstraint(0, 1, ""); for (int j = 0; j < numTasks; ++j) { constraint.setCoefficient(x[i][j], 1); } } // Each task is assigned to exactly one worker. for (int j = 0; j < numTasks; ++j) { MPConstraint constraint = solver.makeConstraint(1, 1, ""); for (int i = 0; i < numWorkers; ++i) { constraint.setCoefficient(x[i][j], 1); } }
C#
// Each worker is assigned to at most one task. for (int i = 0; i < numWorkers; ++i) { Constraint constraint = solver.MakeConstraint(0, 1, ""); for (int j = 0; j < numTasks; ++j) { constraint.SetCoefficient(x[i, j], 1); } } // Each task is assigned to exactly one worker. for (int j = 0; j < numTasks; ++j) { Constraint constraint = solver.MakeConstraint(1, 1, ""); for (int i = 0; i < numWorkers; ++i) { constraint.SetCoefficient(x[i, j], 1); } }
创建目标函数
以下代码为问题创建目标函数。
Python
objective_terms = [] for i in range(num_workers): for j in range(num_tasks): objective_terms.append(costs[i][j] * x[i, j]) solver.Minimize(solver.Sum(objective_terms))
C++
MPObjective* const objective = solver->MutableObjective(); for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { objective->SetCoefficient(x[i][j], costs[i][j]); } } objective->SetMinimization();
Java
MPObjective objective = solver.objective(); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { objective.setCoefficient(x[i][j], costs[i][j]); } } objective.setMinimization();
C#
Objective objective = solver.Objective(); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { objective.SetCoefficient(x[i, j], costs[i, j]); } } objective.SetMinimization();
目标函数的值是分配给所有变量的 求解器求出的值为 1。
调用求解器
以下代码会调用求解器。
Python
print(f"Solving with {solver.SolverVersion()}") status = solver.Solve()
C++
const MPSolver::ResultStatus result_status = solver->Solve();
Java
MPSolver.ResultStatus resultStatus = solver.solve();
C#
Solver.ResultStatus resultStatus = solver.Solve();
输出解决方案
以下代码会输出该问题的解决方案。
Python
if status == pywraplp.Solver.OPTIMAL or status == pywraplp.Solver.FEASIBLE: print(f"Total cost = {solver.Objective().Value()}\n") for i in range(num_workers): for j in range(num_tasks): # Test if x[i,j] is 1 (with tolerance for floating point arithmetic). if x[i, j].solution_value() > 0.5: print(f"Worker {i} assigned to task {j}." + f" Cost: {costs[i][j]}") else: print("No solution found.")
C++
// Check that the problem has a feasible solution. if (result_status != MPSolver::OPTIMAL && result_status != MPSolver::FEASIBLE) { LOG(FATAL) << "No solution found."; } LOG(INFO) << "Total cost = " << objective->Value() << "\n\n"; for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { // Test if x[i][j] is 0 or 1 (with tolerance for floating point // arithmetic). if (x[i][j]->solution_value() > 0.5) { LOG(INFO) << "Worker " << i << " assigned to task " << j << ". Cost = " << costs[i][j]; } } }
Java
// Check that the problem has a feasible solution. if (resultStatus == MPSolver.ResultStatus.OPTIMAL || resultStatus == MPSolver.ResultStatus.FEASIBLE) { System.out.println("Total cost: " + objective.value() + "\n"); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { // Test if x[i][j] is 0 or 1 (with tolerance for floating point // arithmetic). if (x[i][j].solutionValue() > 0.5) { System.out.println( "Worker " + i + " assigned to task " + j + ". Cost = " + costs[i][j]); } } } } else { System.err.println("No solution found."); }
C#
// Check that the problem has a feasible solution. if (resultStatus == Solver.ResultStatus.OPTIMAL || resultStatus == Solver.ResultStatus.FEASIBLE) { Console.WriteLine($"Total cost: {solver.Objective().Value()}\n"); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { // Test if x[i, j] is 0 or 1 (with tolerance for floating point // arithmetic). if (x[i, j].SolutionValue() > 0.5) { Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}"); } } } } else { Console.WriteLine("No solution found."); }
以下是程序的输出。
Total cost = 265.0 Worker 0 assigned to task 3. Cost = 70 Worker 1 assigned to task 2. Cost = 55 Worker 2 assigned to task 1. Cost = 95 Worker 3 assigned to task 0. Cost = 45
完成计划
以下是 MIP 解决方案的完整程序。
Python
from ortools.linear_solver import pywraplp def main(): # Data costs = [ [90, 80, 75, 70], [35, 85, 55, 65], [125, 95, 90, 95], [45, 110, 95, 115], [50, 100, 90, 100], ] num_workers = len(costs) num_tasks = len(costs[0]) # Solver # Create the mip solver with the SCIP backend. solver = pywraplp.Solver.CreateSolver("SCIP") if not solver: return # Variables # x[i, j] is an array of 0-1 variables, which will be 1 # if worker i is assigned to task j. x = {} for i in range(num_workers): for j in range(num_tasks): x[i, j] = solver.IntVar(0, 1, "") # Constraints # Each worker is assigned to at most 1 task. for i in range(num_workers): solver.Add(solver.Sum([x[i, j] for j in range(num_tasks)]) <= 1) # Each task is assigned to exactly one worker. for j in range(num_tasks): solver.Add(solver.Sum([x[i, j] for i in range(num_workers)]) == 1) # Objective objective_terms = [] for i in range(num_workers): for j in range(num_tasks): objective_terms.append(costs[i][j] * x[i, j]) solver.Minimize(solver.Sum(objective_terms)) # Solve print(f"Solving with {solver.SolverVersion()}") status = solver.Solve() # Print solution. if status == pywraplp.Solver.OPTIMAL or status == pywraplp.Solver.FEASIBLE: print(f"Total cost = {solver.Objective().Value()}\n") for i in range(num_workers): for j in range(num_tasks): # Test if x[i,j] is 1 (with tolerance for floating point arithmetic). if x[i, j].solution_value() > 0.5: print(f"Worker {i} assigned to task {j}." + f" Cost: {costs[i][j]}") else: print("No solution found.") if __name__ == "__main__": main()
C++
#include <memory> #include <vector> #include "ortools/base/logging.h" #include "ortools/linear_solver/linear_solver.h" namespace operations_research { void AssignmentMip() { // Data const std::vector<std::vector<double>> costs{ {90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95}, {45, 110, 95, 115}, {50, 100, 90, 100}, }; const int num_workers = costs.size(); const int num_tasks = costs[0].size(); // Solver // Create the mip solver with the SCIP backend. std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP")); if (!solver) { LOG(WARNING) << "SCIP solver unavailable."; return; } // Variables // x[i][j] is an array of 0-1 variables, which will be 1 // if worker i is assigned to task j. std::vector<std::vector<const MPVariable*>> x( num_workers, std::vector<const MPVariable*>(num_tasks)); for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { x[i][j] = solver->MakeIntVar(0, 1, ""); } } // Constraints // Each worker is assigned to at most one task. for (int i = 0; i < num_workers; ++i) { LinearExpr worker_sum; for (int j = 0; j < num_tasks; ++j) { worker_sum += x[i][j]; } solver->MakeRowConstraint(worker_sum <= 1.0); } // Each task is assigned to exactly one worker. for (int j = 0; j < num_tasks; ++j) { LinearExpr task_sum; for (int i = 0; i < num_workers; ++i) { task_sum += x[i][j]; } solver->MakeRowConstraint(task_sum == 1.0); } // Objective. MPObjective* const objective = solver->MutableObjective(); for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { objective->SetCoefficient(x[i][j], costs[i][j]); } } objective->SetMinimization(); // Solve const MPSolver::ResultStatus result_status = solver->Solve(); // Print solution. // Check that the problem has a feasible solution. if (result_status != MPSolver::OPTIMAL && result_status != MPSolver::FEASIBLE) { LOG(FATAL) << "No solution found."; } LOG(INFO) << "Total cost = " << objective->Value() << "\n\n"; for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { // Test if x[i][j] is 0 or 1 (with tolerance for floating point // arithmetic). if (x[i][j]->solution_value() > 0.5) { LOG(INFO) << "Worker " << i << " assigned to task " << j << ". Cost = " << costs[i][j]; } } } } } // namespace operations_research int main(int argc, char** argv) { operations_research::AssignmentMip(); return EXIT_SUCCESS; }
Java
package com.google.ortools.linearsolver.samples; import com.google.ortools.Loader; import com.google.ortools.linearsolver.MPConstraint; import com.google.ortools.linearsolver.MPObjective; import com.google.ortools.linearsolver.MPSolver; import com.google.ortools.linearsolver.MPVariable; /** MIP example that solves an assignment problem. */ public class AssignmentMip { public static void main(String[] args) { Loader.loadNativeLibraries(); // Data double[][] costs = { {90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95}, {45, 110, 95, 115}, {50, 100, 90, 100}, }; int numWorkers = costs.length; int numTasks = costs[0].length; // Solver // Create the linear solver with the SCIP backend. MPSolver solver = MPSolver.createSolver("SCIP"); if (solver == null) { System.out.println("Could not create solver SCIP"); return; } // Variables // x[i][j] is an array of 0-1 variables, which will be 1 // if worker i is assigned to task j. MPVariable[][] x = new MPVariable[numWorkers][numTasks]; for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { x[i][j] = solver.makeIntVar(0, 1, ""); } } // Constraints // Each worker is assigned to at most one task. for (int i = 0; i < numWorkers; ++i) { MPConstraint constraint = solver.makeConstraint(0, 1, ""); for (int j = 0; j < numTasks; ++j) { constraint.setCoefficient(x[i][j], 1); } } // Each task is assigned to exactly one worker. for (int j = 0; j < numTasks; ++j) { MPConstraint constraint = solver.makeConstraint(1, 1, ""); for (int i = 0; i < numWorkers; ++i) { constraint.setCoefficient(x[i][j], 1); } } // Objective MPObjective objective = solver.objective(); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { objective.setCoefficient(x[i][j], costs[i][j]); } } objective.setMinimization(); // Solve MPSolver.ResultStatus resultStatus = solver.solve(); // Print solution. // Check that the problem has a feasible solution. if (resultStatus == MPSolver.ResultStatus.OPTIMAL || resultStatus == MPSolver.ResultStatus.FEASIBLE) { System.out.println("Total cost: " + objective.value() + "\n"); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { // Test if x[i][j] is 0 or 1 (with tolerance for floating point // arithmetic). if (x[i][j].solutionValue() > 0.5) { System.out.println( "Worker " + i + " assigned to task " + j + ". Cost = " + costs[i][j]); } } } } else { System.err.println("No solution found."); } } private AssignmentMip() {} }
C#
using System; using Google.OrTools.LinearSolver; public class AssignmentMip { static void Main() { // Data. int[,] costs = { { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 }, }; int numWorkers = costs.GetLength(0); int numTasks = costs.GetLength(1); // Solver. Solver solver = Solver.CreateSolver("SCIP"); if (solver is null) { return; } // Variables. // x[i, j] is an array of 0-1 variables, which will be 1 // if worker i is assigned to task j. Variable[,] x = new Variable[numWorkers, numTasks]; for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { x[i, j] = solver.MakeIntVar(0, 1, $"worker_{i}_task_{j}"); } } // Constraints // Each worker is assigned to at most one task. for (int i = 0; i < numWorkers; ++i) { Constraint constraint = solver.MakeConstraint(0, 1, ""); for (int j = 0; j < numTasks; ++j) { constraint.SetCoefficient(x[i, j], 1); } } // Each task is assigned to exactly one worker. for (int j = 0; j < numTasks; ++j) { Constraint constraint = solver.MakeConstraint(1, 1, ""); for (int i = 0; i < numWorkers; ++i) { constraint.SetCoefficient(x[i, j], 1); } } // Objective Objective objective = solver.Objective(); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { objective.SetCoefficient(x[i, j], costs[i, j]); } } objective.SetMinimization(); // Solve Solver.ResultStatus resultStatus = solver.Solve(); // Print solution. // Check that the problem has a feasible solution. if (resultStatus == Solver.ResultStatus.OPTIMAL || resultStatus == Solver.ResultStatus.FEASIBLE) { Console.WriteLine($"Total cost: {solver.Objective().Value()}\n"); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { // Test if x[i, j] is 0 or 1 (with tolerance for floating point // arithmetic). if (x[i, j].SolutionValue() > 0.5) { Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}"); } } } } else { Console.WriteLine("No solution found."); } } }
CP SAT 解决方案
下文介绍如何使用 CP-SAT 求解器求解数学题。
导入库
以下代码会导入所需的库。
Python
import io import pandas as pd from ortools.sat.python import cp_model
C++
#include <stdlib.h> #include <vector> #include "ortools/base/logging.h" #include "ortools/sat/cp_model.h" #include "ortools/sat/cp_model.pb.h" #include "ortools/sat/cp_model_solver.h"
Java
import com.google.ortools.Loader; import com.google.ortools.sat.CpModel; import com.google.ortools.sat.CpSolver; import com.google.ortools.sat.CpSolverStatus; import com.google.ortools.sat.LinearExpr; import com.google.ortools.sat.LinearExprBuilder; import com.google.ortools.sat.Literal; import java.util.ArrayList; import java.util.List; import java.util.stream.IntStream;
C#
using System; using System.Collections.Generic; using Google.OrTools.Sat;
声明模型
以下代码声明了 CP-SAT 模型。
Python
model = cp_model.CpModel()
C++
CpModelBuilder cp_model;
Java
CpModel model = new CpModel();
C#
CpModel model = new CpModel();
创建数据
以下代码用于设置问题的数据。
Python
data_str = """ worker task cost w1 t1 90 w1 t2 80 w1 t3 75 w1 t4 70 w2 t1 35 w2 t2 85 w2 t3 55 w2 t4 65 w3 t1 125 w3 t2 95 w3 t3 90 w3 t4 95 w4 t1 45 w4 t2 110 w4 t3 95 w4 t4 115 w5 t1 50 w5 t2 110 w5 t3 90 w5 t4 100 """ data = pd.read_table(io.StringIO(data_str), sep=r"\s+")
C++
const std::vector<std::vector<int>> costs{ {90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95}, {45, 110, 95, 115}, {50, 100, 90, 100}, }; const int num_workers = static_cast<int>(costs.size()); const int num_tasks = static_cast<int>(costs[0].size());
Java
int[][] costs = { {90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95}, {45, 110, 95, 115}, {50, 100, 90, 100}, }; final int numWorkers = costs.length; final int numTasks = costs[0].length; final int[] allWorkers = IntStream.range(0, numWorkers).toArray(); final int[] allTasks = IntStream.range(0, numTasks).toArray();
C#
int[,] costs = { { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 }, }; int numWorkers = costs.GetLength(0); int numTasks = costs.GetLength(1);
costs
数组与费用表相对应
用于向任务分配工作器,如上所示。
创建变量
以下代码会为问题创建二进制整数变量。
Python
x = model.new_bool_var_series(name="x", index=data.index)
C++
// x[i][j] is an array of Boolean variables. x[i][j] is true // if worker i is assigned to task j. std::vector<std::vector<BoolVar>> x(num_workers, std::vector<BoolVar>(num_tasks)); for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { x[i][j] = cp_model.NewBoolVar(); } }
Java
Literal[][] x = new Literal[numWorkers][numTasks]; for (int worker : allWorkers) { for (int task : allTasks) { x[worker][task] = model.newBoolVar("x[" + worker + "," + task + "]"); } }
C#
BoolVar[,] x = new BoolVar[numWorkers, numTasks]; // Variables in a 1-dim array. for (int worker = 0; worker < numWorkers; ++worker) { for (int task = 0; task < numTasks; ++task) { x[worker, task] = model.NewBoolVar($"worker_{worker}_task_{task}"); } }
创建限制条件
以下代码为问题创建了约束条件。
Python
# Each worker is assigned to at most one task. for unused_name, tasks in data.groupby("worker"): model.add_at_most_one(x[tasks.index]) # Each task is assigned to exactly one worker. for unused_name, workers in data.groupby("task"): model.add_exactly_one(x[workers.index])
C++
// Each worker is assigned to at most one task. for (int i = 0; i < num_workers; ++i) { cp_model.AddAtMostOne(x[i]); } // Each task is assigned to exactly one worker. for (int j = 0; j < num_tasks; ++j) { std::vector<BoolVar> tasks; for (int i = 0; i < num_workers; ++i) { tasks.push_back(x[i][j]); } cp_model.AddExactlyOne(tasks); }
Java
// Each worker is assigned to at most one task. for (int worker : allWorkers) { List<Literal> tasks = new ArrayList<>(); for (int task : allTasks) { tasks.add(x[worker][task]); } model.addAtMostOne(tasks); } // Each task is assigned to exactly one worker. for (int task : allTasks) { List<Literal> workers = new ArrayList<>(); for (int worker : allWorkers) { workers.add(x[worker][task]); } model.addExactlyOne(workers); }
C#
// Each worker is assigned to at most one task. for (int worker = 0; worker < numWorkers; ++worker) { List<ILiteral> tasks = new List<ILiteral>(); for (int task = 0; task < numTasks; ++task) { tasks.Add(x[worker, task]); } model.AddAtMostOne(tasks); } // Each task is assigned to exactly one worker. for (int task = 0; task < numTasks; ++task) { List<ILiteral> workers = new List<ILiteral>(); for (int worker = 0; worker < numWorkers; ++worker) { workers.Add(x[worker, task]); } model.AddExactlyOne(workers); }
创建目标函数
以下代码为问题创建目标函数。
Python
model.minimize(data.cost.dot(x))
C++
LinearExpr total_cost; for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { total_cost += x[i][j] * costs[i][j]; } } cp_model.Minimize(total_cost);
Java
LinearExprBuilder obj = LinearExpr.newBuilder(); for (int worker : allWorkers) { for (int task : allTasks) { obj.addTerm(x[worker][task], costs[worker][task]); } } model.minimize(obj);
C#
LinearExprBuilder obj = LinearExpr.NewBuilder(); for (int worker = 0; worker < numWorkers; ++worker) { for (int task = 0; task < numTasks; ++task) { obj.AddTerm((IntVar)x[worker, task], costs[worker, task]); } } model.Minimize(obj);
目标函数的值是分配给所有变量的 求解器求出的值为 1。
调用求解器
以下代码会调用求解器。
Python
solver = cp_model.CpSolver() status = solver.solve(model)
C++
const CpSolverResponse response = Solve(cp_model.Build());
Java
CpSolver solver = new CpSolver(); CpSolverStatus status = solver.solve(model);
C#
CpSolver solver = new CpSolver(); CpSolverStatus status = solver.Solve(model); Console.WriteLine($"Solve status: {status}");
输出解决方案
以下代码会输出该问题的解决方案。
Python
if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE: print(f"Total cost = {solver.objective_value}\n") selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index] for unused_index, row in selected.iterrows(): print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}") elif status == cp_model.INFEASIBLE: print("No solution found") else: print("Something is wrong, check the status and the log of the solve")
C++
if (response.status() == CpSolverStatus::INFEASIBLE) { LOG(FATAL) << "No solution found."; } LOG(INFO) << "Total cost: " << response.objective_value(); LOG(INFO); for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { if (SolutionBooleanValue(response, x[i][j])) { LOG(INFO) << "Task " << i << " assigned to worker " << j << ". Cost: " << costs[i][j]; } } }
Java
// Check that the problem has a feasible solution. if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) { System.out.println("Total cost: " + solver.objectiveValue() + "\n"); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { if (solver.booleanValue(x[i][j])) { System.out.println( "Worker " + i + " assigned to task " + j + ". Cost: " + costs[i][j]); } } } } else { System.err.println("No solution found."); }
C#
// Check that the problem has a feasible solution. if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible) { Console.WriteLine($"Total cost: {solver.ObjectiveValue}\n"); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { if (solver.Value(x[i, j]) > 0.5) { Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}"); } } } } else { Console.WriteLine("No solution found."); }
以下是程序的输出。
Total cost = 265 Worker 0 assigned to task 3 Cost = 70 Worker 1 assigned to task 2 Cost = 55 Worker 2 assigned to task 1 Cost = 95 Worker 3 assigned to task 0 Cost = 45
完成计划
以下是 CP-SAT 解决方案的完整计划。
Python
import io import pandas as pd from ortools.sat.python import cp_model def main() -> None: # Data data_str = """ worker task cost w1 t1 90 w1 t2 80 w1 t3 75 w1 t4 70 w2 t1 35 w2 t2 85 w2 t3 55 w2 t4 65 w3 t1 125 w3 t2 95 w3 t3 90 w3 t4 95 w4 t1 45 w4 t2 110 w4 t3 95 w4 t4 115 w5 t1 50 w5 t2 110 w5 t3 90 w5 t4 100 """ data = pd.read_table(io.StringIO(data_str), sep=r"\s+") # Model model = cp_model.CpModel() # Variables x = model.new_bool_var_series(name="x", index=data.index) # Constraints # Each worker is assigned to at most one task. for unused_name, tasks in data.groupby("worker"): model.add_at_most_one(x[tasks.index]) # Each task is assigned to exactly one worker. for unused_name, workers in data.groupby("task"): model.add_exactly_one(x[workers.index]) # Objective model.minimize(data.cost.dot(x)) # Solve solver = cp_model.CpSolver() status = solver.solve(model) # Print solution. if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE: print(f"Total cost = {solver.objective_value}\n") selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index] for unused_index, row in selected.iterrows(): print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}") elif status == cp_model.INFEASIBLE: print("No solution found") else: print("Something is wrong, check the status and the log of the solve") if __name__ == "__main__": main()
C++
#include <stdlib.h> #include <vector> #include "ortools/base/logging.h" #include "ortools/sat/cp_model.h" #include "ortools/sat/cp_model.pb.h" #include "ortools/sat/cp_model_solver.h" namespace operations_research { namespace sat { void IntegerProgrammingExample() { // Data const std::vector<std::vector<int>> costs{ {90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95}, {45, 110, 95, 115}, {50, 100, 90, 100}, }; const int num_workers = static_cast<int>(costs.size()); const int num_tasks = static_cast<int>(costs[0].size()); // Model CpModelBuilder cp_model; // Variables // x[i][j] is an array of Boolean variables. x[i][j] is true // if worker i is assigned to task j. std::vector<std::vector<BoolVar>> x(num_workers, std::vector<BoolVar>(num_tasks)); for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { x[i][j] = cp_model.NewBoolVar(); } } // Constraints // Each worker is assigned to at most one task. for (int i = 0; i < num_workers; ++i) { cp_model.AddAtMostOne(x[i]); } // Each task is assigned to exactly one worker. for (int j = 0; j < num_tasks; ++j) { std::vector<BoolVar> tasks; for (int i = 0; i < num_workers; ++i) { tasks.push_back(x[i][j]); } cp_model.AddExactlyOne(tasks); } // Objective LinearExpr total_cost; for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { total_cost += x[i][j] * costs[i][j]; } } cp_model.Minimize(total_cost); // Solve const CpSolverResponse response = Solve(cp_model.Build()); // Print solution. if (response.status() == CpSolverStatus::INFEASIBLE) { LOG(FATAL) << "No solution found."; } LOG(INFO) << "Total cost: " << response.objective_value(); LOG(INFO); for (int i = 0; i < num_workers; ++i) { for (int j = 0; j < num_tasks; ++j) { if (SolutionBooleanValue(response, x[i][j])) { LOG(INFO) << "Task " << i << " assigned to worker " << j << ". Cost: " << costs[i][j]; } } } } } // namespace sat } // namespace operations_research int main(int argc, char** argv) { operations_research::sat::IntegerProgrammingExample(); return EXIT_SUCCESS; }
Java
package com.google.ortools.sat.samples; import com.google.ortools.Loader; import com.google.ortools.sat.CpModel; import com.google.ortools.sat.CpSolver; import com.google.ortools.sat.CpSolverStatus; import com.google.ortools.sat.LinearExpr; import com.google.ortools.sat.LinearExprBuilder; import com.google.ortools.sat.Literal; import java.util.ArrayList; import java.util.List; import java.util.stream.IntStream; /** Assignment problem. */ public class AssignmentSat { public static void main(String[] args) { Loader.loadNativeLibraries(); // Data int[][] costs = { {90, 80, 75, 70}, {35, 85, 55, 65}, {125, 95, 90, 95}, {45, 110, 95, 115}, {50, 100, 90, 100}, }; final int numWorkers = costs.length; final int numTasks = costs[0].length; final int[] allWorkers = IntStream.range(0, numWorkers).toArray(); final int[] allTasks = IntStream.range(0, numTasks).toArray(); // Model CpModel model = new CpModel(); // Variables Literal[][] x = new Literal[numWorkers][numTasks]; for (int worker : allWorkers) { for (int task : allTasks) { x[worker][task] = model.newBoolVar("x[" + worker + "," + task + "]"); } } // Constraints // Each worker is assigned to at most one task. for (int worker : allWorkers) { List<Literal> tasks = new ArrayList<>(); for (int task : allTasks) { tasks.add(x[worker][task]); } model.addAtMostOne(tasks); } // Each task is assigned to exactly one worker. for (int task : allTasks) { List<Literal> workers = new ArrayList<>(); for (int worker : allWorkers) { workers.add(x[worker][task]); } model.addExactlyOne(workers); } // Objective LinearExprBuilder obj = LinearExpr.newBuilder(); for (int worker : allWorkers) { for (int task : allTasks) { obj.addTerm(x[worker][task], costs[worker][task]); } } model.minimize(obj); // Solve CpSolver solver = new CpSolver(); CpSolverStatus status = solver.solve(model); // Print solution. // Check that the problem has a feasible solution. if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) { System.out.println("Total cost: " + solver.objectiveValue() + "\n"); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { if (solver.booleanValue(x[i][j])) { System.out.println( "Worker " + i + " assigned to task " + j + ". Cost: " + costs[i][j]); } } } } else { System.err.println("No solution found."); } } private AssignmentSat() {} }
C#
using System; using System.Collections.Generic; using Google.OrTools.Sat; public class AssignmentSat { public static void Main(String[] args) { // Data. int[,] costs = { { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 }, }; int numWorkers = costs.GetLength(0); int numTasks = costs.GetLength(1); // Model. CpModel model = new CpModel(); // Variables. BoolVar[,] x = new BoolVar[numWorkers, numTasks]; // Variables in a 1-dim array. for (int worker = 0; worker < numWorkers; ++worker) { for (int task = 0; task < numTasks; ++task) { x[worker, task] = model.NewBoolVar($"worker_{worker}_task_{task}"); } } // Constraints // Each worker is assigned to at most one task. for (int worker = 0; worker < numWorkers; ++worker) { List<ILiteral> tasks = new List<ILiteral>(); for (int task = 0; task < numTasks; ++task) { tasks.Add(x[worker, task]); } model.AddAtMostOne(tasks); } // Each task is assigned to exactly one worker. for (int task = 0; task < numTasks; ++task) { List<ILiteral> workers = new List<ILiteral>(); for (int worker = 0; worker < numWorkers; ++worker) { workers.Add(x[worker, task]); } model.AddExactlyOne(workers); } // Objective LinearExprBuilder obj = LinearExpr.NewBuilder(); for (int worker = 0; worker < numWorkers; ++worker) { for (int task = 0; task < numTasks; ++task) { obj.AddTerm((IntVar)x[worker, task], costs[worker, task]); } } model.Minimize(obj); // Solve CpSolver solver = new CpSolver(); CpSolverStatus status = solver.Solve(model); Console.WriteLine($"Solve status: {status}"); // Print solution. // Check that the problem has a feasible solution. if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible) { Console.WriteLine($"Total cost: {solver.ObjectiveValue}\n"); for (int i = 0; i < numWorkers; ++i) { for (int j = 0; j < numTasks; ++j) { if (solver.Value(x[i, j]) > 0.5) { Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}"); } } } } else { Console.WriteLine("No solution found."); } Console.WriteLine("Statistics"); Console.WriteLine($" - conflicts : {solver.NumConflicts()}"); Console.WriteLine($" - branches : {solver.NumBranches()}"); Console.WriteLine($" - wall time : {solver.WallTime()}s"); } }